Find $\lim_{x\to -4}\dfrac{x^2-15}{x^2-16}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{1}{8}$ (Choice B) B $\dfrac{1}{8}$ (Choice C) C $\dfrac{19}{20}$ (Choice D) D The limit doesn't exist
Answer: Let's try to find the limit using direct substitution. $\begin{aligned} \lim_{x\to -4}\dfrac{x^2-15}{x^2-16}&=\dfrac{(-4)^2-15}{(-4)^2-16} \\\\ &=\dfrac{1}{0} \end{aligned}$ Our expression evaluates to a nonzero number over zero. In such cases, the limit doesn't exist. In conclusion, $\lim_{x\to -4}\dfrac{x^2-15}{x^2-16}$ doesn't exist.